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Just out of curiosity, does the sum $$\sum_{n=0}^N{1\over\sqrt{n!}}$$ have a closed form for $N<\infty$ or eventually $N\to\infty$ ? I cannot find it anywhere and it does not resemble any function expansion I can now recall.

user155002
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    No. It doesn't. – Lucian Dec 13 '14 at 02:34
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    Approximate value, in case others were wondering: 3.46951 – apnorton Dec 13 '14 at 02:39
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    Perhaps a silly question, but how do you know? Or how can you disprove it? Is this sum somehow important that somebody cares? – user155002 Dec 13 '14 at 02:40
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    Calculate to many decimal places and check at http://isc.carma.newcastle.edu.au/ – Will Jagy Dec 13 '14 at 02:52
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    Cf. to A248761, which is this value minus one with ~100 digits available. As far as I can see, there is no available literature on this sum. – George V. Williams Dec 13 '14 at 02:53
  • When I was a schoolboy, trying to make a tiny discovery, I numerically calculated with high precision similar sums of exotic series with simple written terms, and then compared them with simple expressions obtained from known constants. I did no discovery. :-) – Alex Ravsky Dec 13 '14 at 08:12
  • @Will Jagy Tried and found nothing (nice site btw). – user155002 Dec 13 '14 at 12:07
  • @All Now with a rested mind I do realize that $$\sum_{n=0}^\infty{\alpha^{n+1}\over\sqrt{n!}}|n\rangle$$ is something called a coherent state in physics (unnormalized) so it is certainly important. Nonetheless, this does not answer my question. – user155002 Dec 13 '14 at 12:08

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